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L-functions: its past and future
Shou-Wu Zhang
Princeton University
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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Sum of powers
Question
For an integer k , is there a formula for
Sk(n) := 1k + 2k + 3k + · · ·+ nk?
S0(n) = 1 + 1 + · · ·+ 1 = n, Definition
S1(n) = 1 + 2 + 3 + · · ·+ n = n(n+1)2 , Pythagoras(550BC )
S2(n) = 12 + 22 + · · ·+ n2 = n(n+1)(2n+1)6 , Archimed(250BC )
S3(n) = 13 + 23 + · · ·+ n3 = n2(n+1)2
4 , Aryabhata(476AD)
Shou-Wu Zhang L-functions: its past and future
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General formula
General formulae were obtained by Fermat, Pascal, Bernoulli etc:For each k , there is a general expression
1k + 2k + 3k + · · ·+ nk = ak1n + ak2n2 + · · ·+ akknk+1, aki ∈ Q
Hua Luogeng (1910-1985)
Shou-Wu Zhang L-functions: its past and future
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General formula
General formulae were obtained by Fermat, Pascal, Bernoulli etc:
For each k , there is a general expression
1k + 2k + 3k + · · ·+ nk = ak1n + ak2n2 + · · ·+ akknk+1, aki ∈ Q
Hua Luogeng (1910-1985)
Shou-Wu Zhang L-functions: its past and future
![Page 10: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/10.jpg)
General formula
General formulae were obtained by Fermat, Pascal, Bernoulli etc:For each k , there is a general expression
1k + 2k + 3k + · · ·+ nk = ak1n + ak2n2 + · · ·+ akknk+1, aki ∈ Q
Hua Luogeng (1910-1985)
Shou-Wu Zhang L-functions: its past and future
![Page 11: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/11.jpg)
General formula
General formulae were obtained by Fermat, Pascal, Bernoulli etc:For each k , there is a general expression
1k + 2k + 3k + · · ·+ nk = ak1n + ak2n2 + · · ·+ akknk+1, aki ∈ Q
Hua Luogeng (1910-1985)
Shou-Wu Zhang L-functions: its past and future
![Page 12: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/12.jpg)
General formula
General formulae were obtained by Fermat, Pascal, Bernoulli etc:For each k , there is a general expression
1k + 2k + 3k + · · ·+ nk = ak1n + ak2n2 + · · ·+ akknk+1, aki ∈ Q
Hua Luogeng (1910-1985)
Shou-Wu Zhang L-functions: its past and future
![Page 13: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/13.jpg)
Art of Conjecturing
Jacobi Bernoulli(1655–1705)
Bernoulli numbers Bj :z
ez−1 =∑∞
j=0 Bjz j
j! .
Sk(n) = 1k+1
∑kj=0(−1)jC k+1
j Bjnk+1−j = (B+n)k+1−Bk+1
k+1 .
Shou-Wu Zhang L-functions: its past and future
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Art of Conjecturing
Jacobi Bernoulli(1655–1705)
Bernoulli numbers Bj :z
ez−1 =∑∞
j=0 Bjz j
j! .
Sk(n) = 1k+1
∑kj=0(−1)jC k+1
j Bjnk+1−j = (B+n)k+1−Bk+1
k+1 .
Shou-Wu Zhang L-functions: its past and future
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Art of Conjecturing
Jacobi Bernoulli(1655–1705)
Bernoulli numbers Bj :z
ez−1 =∑∞
j=0 Bjz j
j! .
Sk(n) = 1k+1
∑kj=0(−1)jC k+1
j Bjnk+1−j = (B+n)k+1−Bk+1
k+1 .
Shou-Wu Zhang L-functions: its past and future
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Art of Conjecturing
Jacobi Bernoulli(1655–1705)
Bernoulli numbers Bj :z
ez−1 =∑∞
j=0 Bjz j
j! .
Sk(n) = 1k+1
∑kj=0(−1)jC k+1
j Bjnk+1−j = (B+n)k+1−Bk+1
k+1 .
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1
n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2n
n3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Yang Hui – Pascal triangle
Yang Hui (1238–1298)
n − (n − 1) = 1n2 − (n − 1)2 = −1 + 2nn3−(n−1)3 = 1−3n+3n2
nk − (n − 1)k =ck1 + ck2n + ck3n2 + · · ·
Shou-Wu Zhang L-functions: its past and future
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Sk(n) = ak1n + ak2n2 + · · ·+ akknk+1
nk − (n − 1)k = ck1 + ck2n + ck3n2 + · · ·+ ckknk−1
a11 0 0 0 · · ·a21 a22 0 0 · · ·a31 a32 a33 0 · · ·
......
......
...
=
c11 0 0 0 · · ·c21 c22 0 0 · · ·c31 c32 c33 0 · · ·
......
......
...
−1
Shou-Wu Zhang L-functions: its past and future
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Sk(n) = ak1n + ak2n2 + · · ·+ akknk+1
nk − (n − 1)k = ck1 + ck2n + ck3n2 + · · ·+ ckknk−1
a11 0 0 0 · · ·a21 a22 0 0 · · ·a31 a32 a33 0 · · ·
......
......
...
=
c11 0 0 0 · · ·c21 c22 0 0 · · ·c31 c32 c33 0 · · ·
......
......
...
−1
Shou-Wu Zhang L-functions: its past and future
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Sk(n) = ak1n + ak2n2 + · · ·+ akknk+1
nk − (n − 1)k = ck1 + ck2n + ck3n2 + · · ·+ ckknk−1
a11 0 0 0 · · ·a21 a22 0 0 · · ·a31 a32 a33 0 · · ·
......
......
...
=
c11 0 0 0 · · ·c21 c22 0 0 · · ·c31 c32 c33 0 · · ·
......
......
...
−1
Shou-Wu Zhang L-functions: its past and future
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proof
Set G (z , n) =∑∞
k=0 Sk(n) zk
k! .
Then
G (z , n) =1− enz
e−z − 1=∞∑j=0
Bj(−z)j−1
j!
∞∑i=1
−(nz)i
i !
=∞∑k=0
zk
k!
1
k + 1
k∑j=0
(−1)jC jk+1Bjn
k+1−j .
Shou-Wu Zhang L-functions: its past and future
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proof
Set G (z , n) =∑∞
k=0 Sk(n) zk
k! .Then
G (z , n) =1− enz
e−z − 1=∞∑j=0
Bj(−z)j−1
j!
∞∑i=1
−(nz)i
i !
=∞∑k=0
zk
k!
1
k + 1
k∑j=0
(−1)jC jk+1Bjn
k+1−j .
Shou-Wu Zhang L-functions: its past and future
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Sum of negative powers
How about negative powers
S−k(n) = 1 +1
2k+ · · ·+ 1
nk?
Do these values extend to nice (smooth, analytic, etc) functionS(x)? Amazingly, this is solved by Euler (1707–1783) when he was26 years old.
For any nice function f (x), thesum S(n) =
∑ni=1 f (i) extends
to a nice function S(x).
Shou-Wu Zhang L-functions: its past and future
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Sum of negative powers
How about negative powers
S−k(n) = 1 +1
2k+ · · ·+ 1
nk?
Do these values extend to nice (smooth, analytic, etc) functionS(x)?
Amazingly, this is solved by Euler (1707–1783) when he was26 years old.
For any nice function f (x), thesum S(n) =
∑ni=1 f (i) extends
to a nice function S(x).
Shou-Wu Zhang L-functions: its past and future
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Sum of negative powers
How about negative powers
S−k(n) = 1 +1
2k+ · · ·+ 1
nk?
Do these values extend to nice (smooth, analytic, etc) functionS(x)? Amazingly, this is solved by Euler (1707–1783) when he was26 years old.
For any nice function f (x), thesum S(n) =
∑ni=1 f (i) extends
to a nice function S(x).
Shou-Wu Zhang L-functions: its past and future
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Sum of negative powers
How about negative powers
S−k(n) = 1 +1
2k+ · · ·+ 1
nk?
Do these values extend to nice (smooth, analytic, etc) functionS(x)? Amazingly, this is solved by Euler (1707–1783) when he was26 years old.
For any nice function f (x), thesum S(n) =
∑ni=1 f (i) extends
to a nice function S(x).
Shou-Wu Zhang L-functions: its past and future
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Euler formula
The relation between f (x) and S(x) is given by difference equation:
f (x) = S(x)− S(x − 1).
Using the Taylor expansion
S(x − 1) = S(x)− S ′(x) +1
2S ′′(x) + · · · = e−DS(x), D =
d
dx,
then
f (x) = (1− e−D)S(x) =1− e−D
DDS(x)
Shou-Wu Zhang L-functions: its past and future
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Euler formula
The relation between f (x) and S(x) is given by difference equation:
f (x) = S(x)− S(x − 1).
Using the Taylor expansion
S(x − 1) = S(x)− S ′(x) +1
2S ′′(x) + · · · = e−DS(x), D =
d
dx,
then
f (x) = (1− e−D)S(x) =1− e−D
DDS(x)
Shou-Wu Zhang L-functions: its past and future
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Euler formula
The relation between f (x) and S(x) is given by difference equation:
f (x) = S(x)− S(x − 1).
Using the Taylor expansion
S(x − 1) = S(x)− S ′(x) +1
2S ′′(x) + · · · = e−DS(x), D =
d
dx,
then
f (x) = (1− e−D)S(x) =1− e−D
DDS(x)
Shou-Wu Zhang L-functions: its past and future
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Thus we obtain a differential equation
DS(x) =D
1− e−Df (x) =
∞∑i=0
(−1)iBiD i
i!f (x).
S(x) =
∫f (x)dx +
∞∑i=1
(−1)iBiD i−1
i!f (x).
For f (x) = xk with k ≥ 0 an integer, Euler’s formula recoversBernoulli’s formula.
Shou-Wu Zhang L-functions: its past and future
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Thus we obtain a differential equation
DS(x) =D
1− e−Df (x) =
∞∑i=0
(−1)iBiD i
i!f (x).
S(x) =
∫f (x)dx +
∞∑i=1
(−1)iBiD i−1
i!f (x).
For f (x) = xk with k ≥ 0 an integer, Euler’s formula recoversBernoulli’s formula.
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Thus we obtain a differential equation
DS(x) =D
1− e−Df (x) =
∞∑i=0
(−1)iBiD i
i!f (x).
S(x) =
∫f (x)dx +
∞∑i=1
(−1)iBiD i−1
i!f (x).
For f (x) = xk with k ≥ 0 an integer, Euler’s formula recoversBernoulli’s formula.
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For each k < 0, one needs to determine a constant in the formula:
S−1(n) = γ + log n +1
2n− 1
12n2+ · · · , (γ = 0.57721...)
S−k(n) = S−k(∞) +1
(k − 1)nk−1 +1
2nk+ · · · , (k > 1)
Question
For each real k > 1, how to evaluate the infinite sum:
ζ(k) := S−k(∞) = 1 +1
2k+
1
3k+ · · ·?
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For each k < 0, one needs to determine a constant in the formula:
S−1(n) = γ + log n +1
2n− 1
12n2+ · · · , (γ = 0.57721...)
S−k(n) = S−k(∞) +1
(k − 1)nk−1 +1
2nk+ · · · , (k > 1)
Question
For each real k > 1, how to evaluate the infinite sum:
ζ(k) := S−k(∞) = 1 +1
2k+
1
3k+ · · ·?
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For each k < 0, one needs to determine a constant in the formula:
S−1(n) = γ + log n +1
2n− 1
12n2+ · · · , (γ = 0.57721...)
S−k(n) = S−k(∞) +1
(k − 1)nk−1 +1
2nk+ · · · , (k > 1)
Question
For each real k > 1, how to evaluate the infinite sum:
ζ(k) := S−k(∞) = 1 +1
2k+
1
3k+ · · ·?
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For each k < 0, one needs to determine a constant in the formula:
S−1(n) = γ + log n +1
2n− 1
12n2+ · · · , (γ = 0.57721...)
S−k(n) = S−k(∞) +1
(k − 1)nk−1 +1
2nk+ · · · , (k > 1)
Question
For each real k > 1, how to evaluate the infinite sum:
ζ(k) := S−k(∞) = 1 +1
2k+
1
3k+ · · ·?
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For each k < 0, one needs to determine a constant in the formula:
S−1(n) = γ + log n +1
2n− 1
12n2+ · · · , (γ = 0.57721...)
S−k(n) = S−k(∞) +1
(k − 1)nk−1 +1
2nk+ · · · , (k > 1)
Question
For each real k > 1, how to evaluate the infinite sum:
ζ(k) := S−k(∞) = 1 +1
2k+
1
3k+ · · ·?
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Basel problem
In 1735, Euler made the first major contribution to this problem:he solved the case k = 2 which was called Basel problem:
∞∑n=1
1
n2=π2
6.
He observed the fact sin x = 0 if and only if x = 0,±π,±2π · · ·which leads to a product formula:
sin x = x∞∏k=1
(1− x2
k2π2).
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Basel problem
In 1735, Euler made the first major contribution to this problem:he solved the case k = 2 which was called Basel problem:
∞∑n=1
1
n2=π2
6.
He observed the fact sin x = 0 if and only if x = 0,±π,±2π · · ·which leads to a product formula:
sin x = x∞∏k=1
(1− x2
k2π2).
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Basel problem
In 1735, Euler made the first major contribution to this problem:he solved the case k = 2 which was called Basel problem:
∞∑n=1
1
n2=π2
6.
He observed the fact sin x = 0 if and only if x = 0,±π,±2π · · ·which leads to a product formula:
sin x = x∞∏k=1
(1− x2
k2π2).
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On the other hand, there is also an addition formula:
sin x = x − x3
3!+
x5
5!+ · · · .
Comparing the coefficients of x3 in these two expressions gives
∞∑k=1
1
k2π2=
1
3!.
The argument also applies to symmetric functions of 1/π2k2 andthus gives a formula for ζ(2k). More precisely, two expansions ofcot x = (log sin x)′ will give
ζ(2k) =(−1)k−1B2k(2π)2k
2(2k)!
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On the other hand, there is also an addition formula:
sin x = x − x3
3!+
x5
5!+ · · · .
Comparing the coefficients of x3 in these two expressions gives
∞∑k=1
1
k2π2=
1
3!.
The argument also applies to symmetric functions of 1/π2k2 andthus gives a formula for ζ(2k). More precisely, two expansions ofcot x = (log sin x)′ will give
ζ(2k) =(−1)k−1B2k(2π)2k
2(2k)!
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On the other hand, there is also an addition formula:
sin x = x − x3
3!+
x5
5!+ · · · .
Comparing the coefficients of x3 in these two expressions gives
∞∑k=1
1
k2π2=
1
3!.
The argument also applies to symmetric functions of 1/π2k2 andthus gives a formula for ζ(2k). More precisely, two expansions ofcot x = (log sin x)′ will give
ζ(2k) =(−1)k−1B2k(2π)2k
2(2k)!
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Euler product
In 1737, Euler discovered another amazing property of ζ(s) whenhe was 30 years old:
ζ(s) =∏p
1
1− 1ps. (s > 1)
Its immediate application is to take the expansion of log ζ(s) togive
log ζ(s) = −∑p
log(1− 1
ps) =
∑p
1
ps+ O(1).
Take s = 1 to give ∑ 1
p=∞.
This is a major refinement of the infiniteness of primes proved byEuclid (300 BC).
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Euler product
In 1737, Euler discovered another amazing property of ζ(s) whenhe was 30 years old:
ζ(s) =∏p
1
1− 1ps. (s > 1)
Its immediate application is to take the expansion of log ζ(s) togive
log ζ(s) = −∑p
log(1− 1
ps) =
∑p
1
ps+ O(1).
Take s = 1 to give ∑ 1
p=∞.
This is a major refinement of the infiniteness of primes proved byEuclid (300 BC).
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Euler product
In 1737, Euler discovered another amazing property of ζ(s) whenhe was 30 years old:
ζ(s) =∏p
1
1− 1ps. (s > 1)
Its immediate application is to take the expansion of log ζ(s) togive
log ζ(s) = −∑p
log(1− 1
ps) =
∑p
1
ps+ O(1).
Take s = 1 to give ∑ 1
p=∞.
This is a major refinement of the infiniteness of primes proved byEuclid (300 BC).
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Euler product
In 1737, Euler discovered another amazing property of ζ(s) whenhe was 30 years old:
ζ(s) =∏p
1
1− 1ps. (s > 1)
Its immediate application is to take the expansion of log ζ(s) togive
log ζ(s) = −∑p
log(1− 1
ps) =
∑p
1
ps+ O(1).
Take s = 1 to give ∑ 1
p=∞.
This is a major refinement of the infiniteness of primes proved byEuclid (300 BC).
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Euler product
In 1737, Euler discovered another amazing property of ζ(s) whenhe was 30 years old:
ζ(s) =∏p
1
1− 1ps. (s > 1)
Its immediate application is to take the expansion of log ζ(s) togive
log ζ(s) = −∑p
log(1− 1
ps) =
∑p
1
ps+ O(1).
Take s = 1 to give ∑ 1
p=∞.
This is a major refinement of the infiniteness of primes proved byEuclid (300 BC).
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Euler zeta function for positive reals
Euler extended ζ(s) to reals in (0, 1) by the following methods:
ζ(s)− 2 · 2−sζ(s) = 1− 1
2s+
1
3s− · · · .
Thus define
ζ(s) =1
1− 21−s(1− 1
2s+
1
3s− · · · ), (s > 0).
It has an asymptotic behavior near s = 1:
ζ(s) =1
s − 1+ O(1).
He further extended it to negative of s when the limit exists:
ζ(s) =1
1− 21−slim
x→1−(x − x2
2s+
x3
3s− · · · ).
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Euler zeta function for positive reals
Euler extended ζ(s) to reals in (0, 1) by the following methods:
ζ(s)− 2 · 2−sζ(s) = 1− 1
2s+
1
3s− · · · .
Thus define
ζ(s) =1
1− 21−s(1− 1
2s+
1
3s− · · · ), (s > 0).
It has an asymptotic behavior near s = 1:
ζ(s) =1
s − 1+ O(1).
He further extended it to negative of s when the limit exists:
ζ(s) =1
1− 21−slim
x→1−(x − x2
2s+
x3
3s− · · · ).
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Euler zeta function for positive reals
Euler extended ζ(s) to reals in (0, 1) by the following methods:
ζ(s)− 2 · 2−sζ(s) = 1− 1
2s+
1
3s− · · · .
Thus define
ζ(s) =1
1− 21−s(1− 1
2s+
1
3s− · · · ), (s > 0).
It has an asymptotic behavior near s = 1:
ζ(s) =1
s − 1+ O(1).
He further extended it to negative of s when the limit exists:
ζ(s) =1
1− 21−slim
x→1−(x − x2
2s+
x3
3s− · · · ).
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Euler zeta function for positive reals
Euler extended ζ(s) to reals in (0, 1) by the following methods:
ζ(s)− 2 · 2−sζ(s) = 1− 1
2s+
1
3s− · · · .
Thus define
ζ(s) =1
1− 21−s(1− 1
2s+
1
3s− · · · ), (s > 0).
It has an asymptotic behavior near s = 1:
ζ(s) =1
s − 1+ O(1).
He further extended it to negative of s when the limit exists:
ζ(s) =1
1− 21−slim
x→1−(x − x2
2s+
x3
3s− · · · ).
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Zeta values at negative integers
In 1739, Euler evaluated the values of ζ at negative integers andobtained
ζ(−k) = (−1)kBk+1
k + 1, k < 0
Euler proved his formula by the following expression:
ζ(−k) = (1− 2k+1)
(x
d
dx
)k ∣∣∣x=1
x
1 + x.
In comparison with ζ(2k) = (−1)k−1B2k (2π)2k
2(2k)! , there is a “functional
equation” between ζ(s) and ζ(1− s) for integers.
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Zeta values at negative integers
In 1739, Euler evaluated the values of ζ at negative integers andobtained
ζ(−k) = (−1)kBk+1
k + 1, k < 0
Euler proved his formula by the following expression:
ζ(−k) = (1− 2k+1)
(x
d
dx
)k ∣∣∣x=1
x
1 + x.
In comparison with ζ(2k) = (−1)k−1B2k (2π)2k
2(2k)! , there is a “functional
equation” between ζ(s) and ζ(1− s) for integers.
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Zeta values at negative integers
In 1739, Euler evaluated the values of ζ at negative integers andobtained
ζ(−k) = (−1)kBk+1
k + 1, k < 0
Euler proved his formula by the following expression:
ζ(−k) = (1− 2k+1)
(x
d
dx
)k ∣∣∣x=1
x
1 + x.
In comparison with ζ(2k) = (−1)k−1B2k (2π)2k
2(2k)! , there is a “functional
equation” between ζ(s) and ζ(1− s) for integers.
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Arithmetic progress of primes
In 1837, Dirichlet introduced his L-function to study the arithmeticprogress of primes.
Dirichlet (1805–1859)
For two integers N, a coprime toeach other, there are infinitelymany primes p such that p ≡ amod N.Moreover, ∑
p≡a mod N
1
p=∞.
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Arithmetic progress of primes
In 1837, Dirichlet introduced his L-function to study the arithmeticprogress of primes.
Dirichlet (1805–1859)
For two integers N, a coprime toeach other, there are infinitelymany primes p such that p ≡ amod N.
Moreover, ∑p≡a mod N
1
p=∞.
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Arithmetic progress of primes
In 1837, Dirichlet introduced his L-function to study the arithmeticprogress of primes.
Dirichlet (1805–1859)
For two integers N, a coprime toeach other, there are infinitelymany primes p such that p ≡ amod N.Moreover, ∑
p≡a mod N
1
p=∞.
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Fourier analysis on finite group
Let G be a finite abelian group with uniform probabilistic measure.Then there is an orthogonal decomposition
L2(G ) =∑χ∈G
Cχ, G := Hom(G ,C×).
Thus for any function f on G , there is a Fourier expansion
f =∑χ
〈f , χ〉χ, 〈f , χ〉 =1
#G
∑g∈G
f (g)χ(g).
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Fourier analysis on finite group
Let G be a finite abelian group with uniform probabilistic measure.Then there is an orthogonal decomposition
L2(G ) =∑χ∈G
Cχ, G := Hom(G ,C×).
Thus for any function f on G , there is a Fourier expansion
f =∑χ
〈f , χ〉χ, 〈f , χ〉 =1
#G
∑g∈G
f (g)χ(g).
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Fourier analysis on multiplicative group
Apply to G = (Z/NZ)× and f = δa, we obtain
δa =∑χ
〈δa, χ〉χ.
Consequently,∑p≡a mod N
1
p=∑p
δa(p)
p=∑χ
〈δa, χ〉∑p
χ(p)
p.
When χ = 1,∑
pχ(p)p =∞ by Euler, it suffices to show when
χ 6= 1 ∑p
χ(p)
p6=∞
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Fourier analysis on multiplicative group
Apply to G = (Z/NZ)× and f = δa, we obtain
δa =∑χ
〈δa, χ〉χ.
Consequently,∑p≡a mod N
1
p=∑p
δa(p)
p=∑χ
〈δa, χ〉∑p
χ(p)
p.
When χ = 1,∑
pχ(p)p =∞ by Euler, it suffices to show when
χ 6= 1 ∑p
χ(p)
p6=∞
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Fourier analysis on multiplicative group
Apply to G = (Z/NZ)× and f = δa, we obtain
δa =∑χ
〈δa, χ〉χ.
Consequently,∑p≡a mod N
1
p=∑p
δa(p)
p=∑χ
〈δa, χ〉∑p
χ(p)
p.
When χ = 1,∑
pχ(p)p =∞ by Euler, it suffices to show when
χ 6= 1 ∑p
χ(p)
p6=∞
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Dirichlet L-function
For χ a character of (Z/NZ)×, Dirichlet introduce his function
L(χ, s) =∑n
χ(n)
ns=∏p
1
1− χ(p)ps
, s > 1.
Then
log L(χ, 1) =∑ χ(p)
p+ O(1).
So is suffices to show that L(χ, 1) is finite and nonzero.
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Dirichlet L-function
For χ a character of (Z/NZ)×, Dirichlet introduce his function
L(χ, s) =∑n
χ(n)
ns=∏p
1
1− χ(p)ps
, s > 1.
Then
log L(χ, 1) =∑ χ(p)
p+ O(1).
So is suffices to show that L(χ, 1) is finite and nonzero.
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Dirichlet L-function
For χ a character of (Z/NZ)×, Dirichlet introduce his function
L(χ, s) =∑n
χ(n)
ns=∏p
1
1− χ(p)ps
, s > 1.
Then
log L(χ, 1) =∑ χ(p)
p+ O(1).
So is suffices to show that L(χ, 1) is finite and nonzero.
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Fourier analysis on additive group
Now we apply Fourier analysis to M = Z/NZ.
L2(M) =∑M
Cψ, M := Hom(M,C×).
Extend G ⊂ M, then we have Fourier expansion
χ =∑〈χ, ψ〉ψ, 〈χ, ψ〉 =
1
N
∑m∈M
ψ(m)χ(m).
This implies
L(χ, s) =∑ψ
〈χ, ψ〉L(ψ, s), L(ψ, s) =∑n
ψ(n)
ns
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Dirichlet special value formula
The value L(ψ, 1) is computable:
L(ψ, 1) =∞∑n=1
ψ(1)n
n= − log(1− ψ(1)).
Thus, there is an important special value formula:
L(χ, 1) = − log∏ψ
(1− ψ(1))〈χ,ψ〉.
This would imply that L(χ, 1) is nonzero and finite, and completesthe proof of Dirichlet’s theorem.Example: N = 4, χ : (Z/4Z)× → {±1}, then
L(χ, 1) = 1− 1
3+
1
5· · · =
π
4.
Shou-Wu Zhang L-functions: its past and future
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Dirichlet special value formula
The value L(ψ, 1) is computable:
L(ψ, 1) =∞∑n=1
ψ(1)n
n= − log(1− ψ(1)).
Thus, there is an important special value formula:
L(χ, 1) = − log∏ψ
(1− ψ(1))〈χ,ψ〉.
This would imply that L(χ, 1) is nonzero and finite, and completesthe proof of Dirichlet’s theorem.Example: N = 4, χ : (Z/4Z)× → {±1}, then
L(χ, 1) = 1− 1
3+
1
5· · · =
π
4.
Shou-Wu Zhang L-functions: its past and future
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Dirichlet special value formula
The value L(ψ, 1) is computable:
L(ψ, 1) =∞∑n=1
ψ(1)n
n= − log(1− ψ(1)).
Thus, there is an important special value formula:
L(χ, 1) = − log∏ψ
(1− ψ(1))〈χ,ψ〉.
This would imply that L(χ, 1) is nonzero and finite, and completesthe proof of Dirichlet’s theorem.
Example: N = 4, χ : (Z/4Z)× → {±1}, then
L(χ, 1) = 1− 1
3+
1
5· · · =
π
4.
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Dirichlet special value formula
The value L(ψ, 1) is computable:
L(ψ, 1) =∞∑n=1
ψ(1)n
n= − log(1− ψ(1)).
Thus, there is an important special value formula:
L(χ, 1) = − log∏ψ
(1− ψ(1))〈χ,ψ〉.
This would imply that L(χ, 1) is nonzero and finite, and completesthe proof of Dirichlet’s theorem.Example: N = 4, χ : (Z/4Z)× → {±1}, then
L(χ, 1) = 1− 1
3+
1
5· · · =
π
4.
Shou-Wu Zhang L-functions: its past and future
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Riemann’s memoire
In 1859, in his memoire ”on the number of primes less than a givenquantity”, consider zeta function with complex variable Re(s) > 1:
Riemann (1826–1866)
ζ(s) =∞∑n=1
1
ns=∏p
1
1− 1ps.
Riemann discovered severalextremely important propertiesusing Fourier analysis on R andR×.
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Riemann’s memoire
In 1859, in his memoire ”on the number of primes less than a givenquantity”, consider zeta function with complex variable Re(s) > 1:
Riemann (1826–1866)
ζ(s) =∞∑n=1
1
ns=∏p
1
1− 1ps.
Riemann discovered severalextremely important propertiesusing Fourier analysis on R andR×.
Shou-Wu Zhang L-functions: its past and future
![Page 80: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/80.jpg)
Riemann’s memoire
In 1859, in his memoire ”on the number of primes less than a givenquantity”, consider zeta function with complex variable Re(s) > 1:
Riemann (1826–1866)
ζ(s) =∞∑n=1
1
ns=∏p
1
1− 1ps.
Riemann discovered severalextremely important propertiesusing Fourier analysis on R andR×.
Shou-Wu Zhang L-functions: its past and future
![Page 81: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/81.jpg)
Continuation and functional equation
Using Γ(s) := 〈e−x , x s−1〉 =∫∞0 e−xx s dx
x , one obtains ζ(s) as theMellin transform of a theta function:
ξ(s) := π−s/2Γ(s/2)ζ(s) =
∫ ∞0
f (t)ts/2dt
t, f (t) :=
∞∑n=1
e−πn2t .
Writing∫∞0 =
∫ 10 +
∫∞1 and using Poisson summation formula, we
have
ξ(s) =
∫ ∞1
f (t)(ts/2 + t(1−s)/2)dt
t−(
1
s+
1
1− s
).
This gives the meromorphic continuation and the functionalequation ξ(s) = ξ(1− s). This somehow explains the relationbetween ζ(1− k) and ζ(k) given by Euler.
Shou-Wu Zhang L-functions: its past and future
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Continuation and functional equation
Using Γ(s) := 〈e−x , x s−1〉 =∫∞0 e−xx s dx
x , one obtains ζ(s) as theMellin transform of a theta function:
ξ(s) := π−s/2Γ(s/2)ζ(s) =
∫ ∞0
f (t)ts/2dt
t, f (t) :=
∞∑n=1
e−πn2t .
Writing∫∞0 =
∫ 10 +
∫∞1 and using Poisson summation formula, we
have
ξ(s) =
∫ ∞1
f (t)(ts/2 + t(1−s)/2)dt
t−(
1
s+
1
1− s
).
This gives the meromorphic continuation and the functionalequation ξ(s) = ξ(1− s). This somehow explains the relationbetween ζ(1− k) and ζ(k) given by Euler.
Shou-Wu Zhang L-functions: its past and future
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Continuation and functional equation
Using Γ(s) := 〈e−x , x s−1〉 =∫∞0 e−xx s dx
x , one obtains ζ(s) as theMellin transform of a theta function:
ξ(s) := π−s/2Γ(s/2)ζ(s) =
∫ ∞0
f (t)ts/2dt
t, f (t) :=
∞∑n=1
e−πn2t .
Writing∫∞0 =
∫ 10 +
∫∞1 and using Poisson summation formula, we
have
ξ(s) =
∫ ∞1
f (t)(ts/2 + t(1−s)/2)dt
t−(
1
s+
1
1− s
).
This gives the meromorphic continuation and the functionalequation ξ(s) = ξ(1− s).
This somehow explains the relationbetween ζ(1− k) and ζ(k) given by Euler.
Shou-Wu Zhang L-functions: its past and future
![Page 84: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/84.jpg)
Continuation and functional equation
Using Γ(s) := 〈e−x , x s−1〉 =∫∞0 e−xx s dx
x , one obtains ζ(s) as theMellin transform of a theta function:
ξ(s) := π−s/2Γ(s/2)ζ(s) =
∫ ∞0
f (t)ts/2dt
t, f (t) :=
∞∑n=1
e−πn2t .
Writing∫∞0 =
∫ 10 +
∫∞1 and using Poisson summation formula, we
have
ξ(s) =
∫ ∞1
f (t)(ts/2 + t(1−s)/2)dt
t−(
1
s+
1
1− s
).
This gives the meromorphic continuation and the functionalequation ξ(s) = ξ(1− s). This somehow explains the relationbetween ζ(1− k) and ζ(k) given by Euler.
Shou-Wu Zhang L-functions: its past and future
![Page 85: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/85.jpg)
Riemann hypothesis
There is another product formula of ζ(s) in terms of its zeros (assin x).
Together with the Euler product formula, there is an explicitrelation between the zeros and the distribution of primes.Then Riemann bravely made a hypothesis that all zeros of ξ(x) lieson the line Re(s) = 1/2.Riemann Hypothesis is equivalent to an asymptotic formula for thenumber of primes
π(x) := #{p ≤ x} =
∫ x
2
dt
log t+ Oε(x
12+ε)
which is a conjectural stronger form of the prime number theorem.
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Riemann hypothesis
There is another product formula of ζ(s) in terms of its zeros (assin x).Together with the Euler product formula, there is an explicitrelation between the zeros and the distribution of primes.
Then Riemann bravely made a hypothesis that all zeros of ξ(x) lieson the line Re(s) = 1/2.Riemann Hypothesis is equivalent to an asymptotic formula for thenumber of primes
π(x) := #{p ≤ x} =
∫ x
2
dt
log t+ Oε(x
12+ε)
which is a conjectural stronger form of the prime number theorem.
Shou-Wu Zhang L-functions: its past and future
![Page 87: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/87.jpg)
Riemann hypothesis
There is another product formula of ζ(s) in terms of its zeros (assin x).Together with the Euler product formula, there is an explicitrelation between the zeros and the distribution of primes.Then Riemann bravely made a hypothesis that all zeros of ξ(x) lieson the line Re(s) = 1/2.
Riemann Hypothesis is equivalent to an asymptotic formula for thenumber of primes
π(x) := #{p ≤ x} =
∫ x
2
dt
log t+ Oε(x
12+ε)
which is a conjectural stronger form of the prime number theorem.
Shou-Wu Zhang L-functions: its past and future
![Page 88: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/88.jpg)
Riemann hypothesis
There is another product formula of ζ(s) in terms of its zeros (assin x).Together with the Euler product formula, there is an explicitrelation between the zeros and the distribution of primes.Then Riemann bravely made a hypothesis that all zeros of ξ(x) lieson the line Re(s) = 1/2.Riemann Hypothesis is equivalent to an asymptotic formula for thenumber of primes
π(x) := #{p ≤ x} =
∫ x
2
dt
log t+ Oε(x
12+ε)
which is a conjectural stronger form of the prime number theorem.
Shou-Wu Zhang L-functions: its past and future
![Page 89: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/89.jpg)
Dirichlet series
A general Dirichlet series takes a form
L(s) =∞∑n=1
anns
=∏p
1
Fp(s)an ∈ C.
with Fp a polynomial of p−s with leading coefficient 1 and of afixed degree.
It should have meromorphic continuation to the complex plane,and satisfies a functional equation.How to find a general L-series?
Shou-Wu Zhang L-functions: its past and future
![Page 90: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/90.jpg)
Dirichlet series
A general Dirichlet series takes a form
L(s) =∞∑n=1
anns
=∏p
1
Fp(s)an ∈ C.
with Fp a polynomial of p−s with leading coefficient 1 and of afixed degree.It should have meromorphic continuation to the complex plane,
and satisfies a functional equation.How to find a general L-series?
Shou-Wu Zhang L-functions: its past and future
![Page 91: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/91.jpg)
Dirichlet series
A general Dirichlet series takes a form
L(s) =∞∑n=1
anns
=∏p
1
Fp(s)an ∈ C.
with Fp a polynomial of p−s with leading coefficient 1 and of afixed degree.It should have meromorphic continuation to the complex plane,and satisfies a functional equation.
How to find a general L-series?
Shou-Wu Zhang L-functions: its past and future
![Page 92: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/92.jpg)
Dirichlet series
A general Dirichlet series takes a form
L(s) =∞∑n=1
anns
=∏p
1
Fp(s)an ∈ C.
with Fp a polynomial of p−s with leading coefficient 1 and of afixed degree.It should have meromorphic continuation to the complex plane,and satisfies a functional equation.How to find a general L-series?
Shou-Wu Zhang L-functions: its past and future
![Page 93: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/93.jpg)
Dedekind
In 1877, Dedekind generalized Dirichlet’s work to number fields K .
ζK (s) =∑a
1
N(a)s=∏p
1
1−Np−s.
He has managed to prove the meromorphic continuation and afunctional equation.In particular, ζK (s) has a simple pole at s = 1 with residue given by
2r1(2π)r2hR
w√
dK.
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Dedekind
In 1877, Dedekind generalized Dirichlet’s work to number fields K .
ζK (s) =∑a
1
N(a)s=∏p
1
1−Np−s.
He has managed to prove the meromorphic continuation and afunctional equation.
In particular, ζK (s) has a simple pole at s = 1 with residue given by
2r1(2π)r2hR
w√
dK.
Shou-Wu Zhang L-functions: its past and future
![Page 95: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/95.jpg)
Dedekind
In 1877, Dedekind generalized Dirichlet’s work to number fields K .
ζK (s) =∑a
1
N(a)s=∏p
1
1−Np−s.
He has managed to prove the meromorphic continuation and afunctional equation.In particular, ζK (s) has a simple pole at s = 1 with residue given by
2r1(2π)r2hR
w√
dK.
Shou-Wu Zhang L-functions: its past and future
![Page 96: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/96.jpg)
Hecke and Tate on Grossen character
The work of Dedekind was generalized by Hecke (1918) by takingGrossencharacters: L(s, χ).
Dedekind’s work was reformulated byTate (1950) using harmonic analysis on adeles.There are two different ways to generalize L(s, χ) with χ replacedby automorphic representations and Galois representations.The Langlands program says that these two constructions giveessentially the same set of L-functions.Analytic properties of L(s) can be studied using Fourier analysis onGLn and Mn×n.
Shou-Wu Zhang L-functions: its past and future
![Page 97: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/97.jpg)
Hecke and Tate on Grossen character
The work of Dedekind was generalized by Hecke (1918) by takingGrossencharacters: L(s, χ). Dedekind’s work was reformulated byTate (1950) using harmonic analysis on adeles.
There are two different ways to generalize L(s, χ) with χ replacedby automorphic representations and Galois representations.The Langlands program says that these two constructions giveessentially the same set of L-functions.Analytic properties of L(s) can be studied using Fourier analysis onGLn and Mn×n.
Shou-Wu Zhang L-functions: its past and future
![Page 98: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/98.jpg)
Hecke and Tate on Grossen character
The work of Dedekind was generalized by Hecke (1918) by takingGrossencharacters: L(s, χ). Dedekind’s work was reformulated byTate (1950) using harmonic analysis on adeles.There are two different ways to generalize L(s, χ) with χ replacedby automorphic representations and Galois representations.
The Langlands program says that these two constructions giveessentially the same set of L-functions.Analytic properties of L(s) can be studied using Fourier analysis onGLn and Mn×n.
Shou-Wu Zhang L-functions: its past and future
![Page 99: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/99.jpg)
Hecke and Tate on Grossen character
The work of Dedekind was generalized by Hecke (1918) by takingGrossencharacters: L(s, χ). Dedekind’s work was reformulated byTate (1950) using harmonic analysis on adeles.There are two different ways to generalize L(s, χ) with χ replacedby automorphic representations and Galois representations.The Langlands program says that these two constructions giveessentially the same set of L-functions.
Analytic properties of L(s) can be studied using Fourier analysis onGLn and Mn×n.
Shou-Wu Zhang L-functions: its past and future
![Page 100: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/100.jpg)
Hecke and Tate on Grossen character
The work of Dedekind was generalized by Hecke (1918) by takingGrossencharacters: L(s, χ). Dedekind’s work was reformulated byTate (1950) using harmonic analysis on adeles.There are two different ways to generalize L(s, χ) with χ replacedby automorphic representations and Galois representations.The Langlands program says that these two constructions giveessentially the same set of L-functions.Analytic properties of L(s) can be studied using Fourier analysis onGLn and Mn×n.
Shou-Wu Zhang L-functions: its past and future
![Page 101: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/101.jpg)
Relation with arithmetic geometry
The cohomology of algebraic varieties provides many Galoisrepresentations and thus many L-functions.
A web of conjectures assert that the special values of theseL-functions often give crucial information about the Diophantineproperties of the varieties, such as BSD, Tate, etc.Riemann hypothesis, Langlands program, and special values ofL-series are three major topics of number theory in the 21stcentury.
Shou-Wu Zhang L-functions: its past and future
![Page 102: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/102.jpg)
Relation with arithmetic geometry
The cohomology of algebraic varieties provides many Galoisrepresentations and thus many L-functions.A web of conjectures assert that the special values of theseL-functions often give crucial information about the Diophantineproperties of the varieties, such as BSD, Tate, etc.
Riemann hypothesis, Langlands program, and special values ofL-series are three major topics of number theory in the 21stcentury.
Shou-Wu Zhang L-functions: its past and future
![Page 103: Princeton University - cityu-ias- fileL-functions: its past and future Shou-Wu Zhang Princeton University Shou-Wu Zhang L-functions: its past and future](https://reader030.vdocuments.site/reader030/viewer/2022040102/5e13552482ffa83ac33837e9/html5/thumbnails/103.jpg)
Relation with arithmetic geometry
The cohomology of algebraic varieties provides many Galoisrepresentations and thus many L-functions.A web of conjectures assert that the special values of theseL-functions often give crucial information about the Diophantineproperties of the varieties, such as BSD, Tate, etc.Riemann hypothesis, Langlands program, and special values ofL-series are three major topics of number theory in the 21stcentury.
Shou-Wu Zhang L-functions: its past and future